HDU 2594 Simpsons’ Hidden Talents（s1的前缀是s2的后缀）

本文主要是介绍HDU 2594 Simpsons’ Hidden Talents（s1的前缀是s2的后缀），希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！

1、http://acm.hdu.edu.cn/showproblem.php?pid=2594

2、题目大意：

给定两个字符串s1,s2，求一个最长的子串，该字串是s1的前缀并且是s2的后缀

开始用for循环做的超时，其实直接用KMP模板即可

3、题目：

Simpsons’ Hidden Talents

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1534 Accepted Submission(s): 564

Problem Description

Homer: Marge, I just figured out a way to discover some of the talents we weren’t aware we had.
Marge: Yeah, what is it?
Homer: Take me for example. I want to find out if I have a talent in politics, OK?
Marge: OK.
Homer: So I take some politician’s name, say Clinton, and try to find the length of the longest prefix
in Clinton’s name that is a suffix in my name. That’s how close I am to being a politician like Clinton
Marge: Why on earth choose the longest prefix that is a suffix???
Homer: Well, our talents are deeply hidden within ourselves, Marge.
Marge: So how close are you?
Homer: 0!
Marge: I’m not surprised.
Homer: But you know, you must have some real math talent hidden deep in you.
Marge: How come?
Homer: Riemann and Marjorie gives 3!!!
Marge: Who the heck is Riemann?
Homer: Never mind.
Write a program that, when given strings s1 and s2, finds the longest prefix of s1 that is a suffix of s2.

Input

Input consists of two lines. The first line contains s1 and the second line contains s2. You may assume all letters are in lowercase.

Output

Output consists of a single line that contains the longest string that is a prefix of s1 and a suffix of s2, followed by the length of that prefix. If the longest such string is the empty string, then the output should be 0.
The lengths of s1 and s2 will be at most 50000.

Sample Input

  clinton
homer
riemann
marjorie

Sample Output

  0
rie 3

Source

HDU 2010-05 Programming Contest

Recommend

lcy

4、代码：

#include<stdio.h>
#include<string.h>
#define N 50005
char s1[N];
char s2[N];
int f[N];
void getFail(char *p, int *f)
{
int m=strlen(p);
f[0]=f[1]=0;
for(int i=1; i<m; ++i)
{
int j=f[i];
while(j && p[i]!=p[j])j=f[j];
f[i+1] = p[i]==p[j]?1+j:0;
}
}
int find(char *T,char *p,int *f)//传入的三个参数分别是主字符串，要比较的子模板串，子模板串的模式值数组
{
getFail(p,f);
int n=strlen(T);
int m=strlen(p);
int j=0;
for(int i=0; i<n; ++i)//循环主串中的字符
{
//如果出现字符不匹配，没必要从头循环，只需要从模式值f[j]查找
while(j && T[i]!=p[j])j=f[j];//可以保证循环到最后，即保证是后缀
if(T[i]==p[j])++j;
}
return j;
}
int main()
{
while(scanf("%s%s",s1,s2)!=EOF)
{
int id=find(s2,s1,f);
if(id==0)
printf("0\n");
else
{
for(int i=0;i<id;i++)
printf("%c",s1[i]);
printf(" %d\n",id);
}
}
return 0;
}
/*
clinton
homer
riemann
marjorie
riem
riemori
*/

这篇关于HDU 2594 Simpsons’ Hidden Talents（s1的前缀是s2的后缀）的文章就介绍到这儿，希望我们推荐的文章对编程师们有所帮助！